The number of elements in the set $\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\}$ equals __________.

$x+2 y+3 z=42$ <br/><br/>$x, y, z \geq 0$ <br/><br/>as <br/><br/>$\begin{array}{ll}z=0 & x+2 y=42 \Rightarrow 22 \text { cases } \\\\ z=1 & x+2 y=39 \Rightarrow 20 \text { cases } \\\\ z=2 & x+2 y=36 \Rightarrow 19 \text { cases } \\\\ z=3 & x+2 y=33 \Rightarrow 17 \text { cases } \\\\ z=4 & x+2 y=30 \Rightarrow 16 \text { cases } \\\\ z=5 & x+2 y=27 \Rightarrow 14 \text { cases }\end{array}$ <br/><br/>$z=6 \quad x+2 y=24 \Rightarrow 13$ cases <br/><br/>$z=7 \quad x+2 y=21 \Rightarrow 11$ cases <br/><br/>$z=8 \quad x+2 y=18 \Rightarrow 10$ cases <br/><br/>$z=9 \quad x+2 y=15 \Rightarrow 8$ cases <br/><br/>$z=10 x+2 y=12 \Rightarrow 7$ cases <br/><br/>$z=11 x+2 y=9 \Rightarrow 5$ cases <br/><br/>$z=12 x+2 y=6 \Rightarrow 4$ cases <br/><br/>$z=13 x+2 y=3 \Rightarrow 2$ cases <br/><br/>$z=14 x+2 y=0 \Rightarrow 1$ case <br/><br/>Total = 169